F(z)=1/3z

Solution for F(z)=1/3z equation:

(F)=1/3F

We move all terms to the left:

(F)-(1/3F)=0


Domain of the equation: 3F)!=0

F!=0/1

F!=0

F∈R


We add all the numbers together, and all the variables

F-(+1/3F)=0

We get rid of parentheses

F-1/3F=0

We multiply all the terms by the denominator


F*3F-1=0

Wy multiply elements

3F^2-1=0

a = 3; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·3·(-1)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*3}=\frac{0-2\sqrt{3}}{6}
=-\frac{2\sqrt{3}}{6}
=-\frac{\sqrt{3}}{3}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*3}=\frac{0+2\sqrt{3}}{6}
=\frac{2\sqrt{3}}{6}
=\frac{\sqrt{3}}{3}
$